| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing two standard methods for finding powers of complex numbers: binomial expansion and De Moivre's theorem. Part (a) requires careful but routine binomial expansion, while part (b) involves standard conversion to modulus-argument form and applying De Moivre's theorem. The verification in (b)(iii) is mechanical. While it requires multiple steps and careful arithmetic, it demands no novel insight—just application of well-practiced techniques from the Further Maths syllabus. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02d Exponential form: re^(i*theta)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | DR |
| Answer | Marks |
|---|---|
| = –163 + 16 i [so a = –163 and b = 16] | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Binomial theorem showing correct pattern and coeffs (could |
| Answer | Marks | Guidance |
|---|---|---|
| (3 + i)2= (2 + 23i) | B1 | or 3 + 23i − 1 |
| (3 + i)3= 8i or (3 + i)4 = −8 + 83i | B1 | Either seen |
| (3 + i)5= –163 + 16i so a = –163 and b = 16 | B1 | If without working, award no marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | (i) |
| Answer | Marks |
|---|---|
| So z = 2(cos /6 + i sin /6) | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.5 | modulus = 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | (ii) |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | (their modulus)5 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | (iii) |
| = –163 + 16 i as before | B1 | |
| B1 | Condone no intermediate working for B2 provided 7(b)(ii) |
Question 7:
7 | (a) | DR
(3 + i)5=(3)5 + 5(3)4i + 10(3)3i2
+ 10(3)2i3 + 53i4 + i5
= –163 + 16 i [so a = –163 and b = 16] | M1
A1
A1 | 1.1a
1.1
1.1 | Binomial theorem showing correct pattern and coeffs (could
use 5C , etc
0
correct expression (must evaluate nC s)
r
Alternative method
(3 + i)2= (2 + 23i) | B1 | or 3 + 23i − 1
(3 + i)3= 8i or (3 + i)4 = −8 + 83i | B1 | Either seen
(3 + i)5= –163 + 16i so a = –163 and b = 16 | B1 | If without working, award no marks
[3]
7 | (b) | (i) | 3 + i| = 2
arg(3 + i) = /6
So z = 2(cos /6 + i sin /6) | B1
B1
B1ft
[3] | 1.1
1.1
2.5 | modulus = 2
arg = /6 or 30
ft their modulus and argument
7 | (b) | (ii) | z5 = 25 (cos 5/6 +i sin 5/6) = 32 (cos 5/6 +i sin 5/6) | B1ft
B1ft
[2] | 1.1
1.1 | (their modulus)5
5 their argument
If fully correct (condone 150) by converting −163 + 16i
then allow SC B2 (but see below)
7 | (b) | (iii) | 32(cos 5/6 + i sin 5/6) [= 32(–3/2 + ½ i)]
= –163 + 16 i as before | B1
B1 | Condone no intermediate working for B2 provided 7(b)(ii)
correct and from using de Moivre
Or 322 = (−163)2 + 162 oe B1, tan−1(−16/163) = 5/6 as in
2nd quadrant B1 (but do not allow if (b)(ii) done by
converting). In this case, if de Moivre used here:
32 = 25 or 532 = 2 SCB1,
and 5 = 5 /6 or 5 / 5 = SCB1
NB if (b)(ii) is not fully correct, award no marks for part (iii)
[2]\begin{enumerate}[label=(\alph*)]
\item By expanding $( \sqrt { 3 } + \mathrm { i } ) ^ { 5 }$, express $z ^ { 5 }$ in the form $\mathrm { a } +$ bi where $a$ and $b$ are real and exact.
\item \begin{enumerate}[label=(\roman*)]
\item Express $z$ in modulus-argument form.
\item Hence find $z ^ { 5 }$ in modulus-argument form.
\item Use this result to verify your answers to part (a).
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2023 Q7 [10]}}